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Proof of Raabe's test

Proving Absolute Convergence of a SeriesRoot test for seriesCorrect method of Proving Raabe's test?Series $sum_{n=2}^infty frac{14^n}{3^{3n+4}(3n+7)}$ Convergence or Divergence Using The Ratio TestIf ${a_n}$ is a positive, nonincreasing sequence such that $sum_{n=1}^infty a_n$ converges, then prove that $lim_{ntoinfty}2^na_{2^n} = 0$Comparison test of series with ln functionIf the series $sum_{n=1}^{infty} na_{n} $ converges, then $sum_{n=1}^{infty} na_{n+1} $ also converges?How to calculate limit for Raabe's test of $sum_{n=1}^infty a_n$ where $a_1=1, a_n=sin a_{n-1}$?Ratio test when checking the convergence of seriesProve that the series $sum_{n=1}^infty frac{e^nn!}{n^n}$ divergesConvergence or divergence of $sum_{n=1}^{infty} ln({1+ n^{-4/3}})$

3

2

$begingroup$

Let $sum a_n$ be a series of non-negative terms and let $$L = lim_{ntoinfty}nleft(1-frac{a_{n+1}}{a_n}right)$$
Prove that the series converges (resp. diverges) if $L > 1$ (resp. $L<1$). I've tried, for example, that when $L<1$, $$nleft(1-frac{a_{n+1}}{a_n}-Lright)= frac{n(a_n-a_{n+1}-La_n)}{a_n}gefrac{n(a_n-a_{n+1}-a_n)}{a_n}=frac{-na_{n+1}}{a_n}$$ and then using epsilons and the sort, but I can't get anywhere. Any tips?

P.S. using Kummer's test doesn't count

calculus sequences-and-series limits

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asked Jan 8 '14 at 13:24

GPerezGPerez

4,27611542

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add a comment | 

3

2

$begingroup$

Let $sum a_n$ be a series of non-negative terms and let $$L = lim_{ntoinfty}nleft(1-frac{a_{n+1}}{a_n}right)$$
Prove that the series converges (resp. diverges) if $L > 1$ (resp. $L<1$). I've tried, for example, that when $L<1$, $$nleft(1-frac{a_{n+1}}{a_n}-Lright)= frac{n(a_n-a_{n+1}-La_n)}{a_n}gefrac{n(a_n-a_{n+1}-a_n)}{a_n}=frac{-na_{n+1}}{a_n}$$ and then using epsilons and the sort, but I can't get anywhere. Any tips?

P.S. using Kummer's test doesn't count

calculus sequences-and-series limits

share|cite|improve this question

asked Jan 8 '14 at 13:24

GPerezGPerez

4,27611542

$endgroup$

add a comment | 

$begingroup$

Let $sum a_n$ be a series of non-negative terms and let $$L = lim_{ntoinfty}nleft(1-frac{a_{n+1}}{a_n}right)$$
Prove that the series converges (resp. diverges) if $L > 1$ (resp. $L<1$). I've tried, for example, that when $L<1$, $$nleft(1-frac{a_{n+1}}{a_n}-Lright)= frac{n(a_n-a_{n+1}-La_n)}{a_n}gefrac{n(a_n-a_{n+1}-a_n)}{a_n}=frac{-na_{n+1}}{a_n}$$ and then using epsilons and the sort, but I can't get anywhere. Any tips?

P.S. using Kummer's test doesn't count

calculus sequences-and-series limits

share|cite|improve this question

asked Jan 8 '14 at 13:24

GPerezGPerez

4,27611542

$endgroup$

share|cite|improve this question

asked Jan 8 '14 at 13:24

GPerezGPerez

4,27611542

share|cite|improve this question

asked Jan 8 '14 at 13:24

GPerezGPerez

4,27611542

asked Jan 8 '14 at 13:24

GPerezGPerez

4,27611542

asked Jan 8 '14 at 13:24

GPerezGPerez

4,27611542

3 Answers
3

active

oldest

votes

5

$begingroup$

If $L>1$, choose $epsilongt 0$, $L-epsilon>1$ then

$$1-frac{L-epsilon}{n} > frac{a_{n+1}}{a_n}$$

Choose $p$ such that $1lt plt L-epsilon$, $sumfrac1{n^p}$ converges. $b_n=frac1{n^p}$, if $n$ big enough, then

$$frac{b_{n+1}}{b_n}=(1-frac{1}{n+1})^p=1-frac{p}{n}+O(frac1{n^2})gt 1-frac{L-epsilon}{n}> frac{a_{n+1}}{a_n}$$

so $sum a_n$ converges

share|cite|improve this answer

answered Jan 8 '14 at 14:23

ziang chenziang chen

4,2711647

$endgroup$

add a comment | 

4

$begingroup$

The series even converges absoute, thus set $a_n:=left|a_nright|$

Now (if $n$ is big enough)
$$L = lim_{ntoinfty}nleft(1-frac{a_{n+1}}{a_n}right) Longleftrightarrow frac{a_{n+1}}{a_n}leqfrac{n-L}{n}$$

This is equivalent to
$$left(L-1right)a_n leq left(n-1right)a_n-na_{n+1}$$

Because of $L>1$ the left side is bigger than zero, so

$$0leq left(n-1right)a_n-na_{n+1}Longleftrightarrow na_{n+1}leq left(n-1right)a_n$$

That means that $a_n$ is decreasing and bounded, so it does converges. Now define the sum

$$sum b_n = sum left(n-1right)a_n-na_{n+1}$$

Which is a telescop-sum and thus it converges

But that implies that the sum over $a_n$
$$sum a_n leq sum left(L-1right)a_n leq sum b_n$$
is convergent as well by the comparison test.

share|cite|improve this answer

edited Jan 8 '14 at 14:49

answered Jan 8 '14 at 14:22

user127.0.0.1user127.0.0.1

6,00162139

$endgroup$

add a comment | 

1

$begingroup$

The following is a new argument for the convergence part. Assume that $L>1$. Choose $varepsilon > 0$ small enough such that
$$L-varepsilon > 1.$$
There exists some $1 ll N=N(varepsilon)$ such that
$$nBig( 1- frac {a_{n+1}}{a_n} Big) > L-varepsilon$$
for any $n geq N$, namely
$$frac {a_{n+1}}{a_n} < 1 - frac{L-varepsilon}n = frac{n-(L-varepsilon)}n$$
for any $n geq N$. Since
$L-varepsilon>1$, one can always choose $alpha >1$ in such a way that
$$frac{n-(L-varepsilon)}n < Big( frac{n-1}n Big)^alpha$$
for any $n gg 1$, say $n geq M > N$. Hence for large $n$, we obtain
$$ a_{n+1} leq Big( frac{n-1}n Big)^alpha a_n leq Big( frac{n-2}n Big)^alpha a_{n-1} leq cdots leq Big( frac{M-1}n Big)^alpha a_M.$$
Hence $sum a_n$ converges by the comparison test.

share|cite|improve this answer

answered Mar 13 at 6:34

QA NgôQA Ngô

5915

$endgroup$

add a comment | 

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5

$begingroup$

If $L>1$, choose $epsilongt 0$, $L-epsilon>1$ then

$$1-frac{L-epsilon}{n} > frac{a_{n+1}}{a_n}$$

Choose $p$ such that $1lt plt L-epsilon$, $sumfrac1{n^p}$ converges. $b_n=frac1{n^p}$, if $n$ big enough, then

$$frac{b_{n+1}}{b_n}=(1-frac{1}{n+1})^p=1-frac{p}{n}+O(frac1{n^2})gt 1-frac{L-epsilon}{n}> frac{a_{n+1}}{a_n}$$

so $sum a_n$ converges

share|cite|improve this answer

answered Jan 8 '14 at 14:23

ziang chenziang chen

4,2711647

$endgroup$

add a comment | 

5

$begingroup$

If $L>1$, choose $epsilongt 0$, $L-epsilon>1$ then

$$1-frac{L-epsilon}{n} > frac{a_{n+1}}{a_n}$$

Choose $p$ such that $1lt plt L-epsilon$, $sumfrac1{n^p}$ converges. $b_n=frac1{n^p}$, if $n$ big enough, then

$$frac{b_{n+1}}{b_n}=(1-frac{1}{n+1})^p=1-frac{p}{n}+O(frac1{n^2})gt 1-frac{L-epsilon}{n}> frac{a_{n+1}}{a_n}$$

so $sum a_n$ converges

share|cite|improve this answer

answered Jan 8 '14 at 14:23

ziang chenziang chen

4,2711647

$endgroup$

add a comment | 

$begingroup$

If $L>1$, choose $epsilongt 0$, $L-epsilon>1$ then

$$1-frac{L-epsilon}{n} > frac{a_{n+1}}{a_n}$$

Choose $p$ such that $1lt plt L-epsilon$, $sumfrac1{n^p}$ converges. $b_n=frac1{n^p}$, if $n$ big enough, then

$$frac{b_{n+1}}{b_n}=(1-frac{1}{n+1})^p=1-frac{p}{n}+O(frac1{n^2})gt 1-frac{L-epsilon}{n}> frac{a_{n+1}}{a_n}$$

so $sum a_n$ converges

share|cite|improve this answer

answered Jan 8 '14 at 14:23

ziang chenziang chen

4,2711647

$endgroup$

share|cite|improve this answer

answered Jan 8 '14 at 14:23

ziang chenziang chen

4,2711647

answered Jan 8 '14 at 14:23

ziang chenziang chen

4,2711647

answered Jan 8 '14 at 14:23

ziang chenziang chen

4,2711647

4

$begingroup$

The series even converges absoute, thus set $a_n:=left|a_nright|$

Now (if $n$ is big enough)
$$L = lim_{ntoinfty}nleft(1-frac{a_{n+1}}{a_n}right) Longleftrightarrow frac{a_{n+1}}{a_n}leqfrac{n-L}{n}$$

This is equivalent to
$$left(L-1right)a_n leq left(n-1right)a_n-na_{n+1}$$

Because of $L>1$ the left side is bigger than zero, so

$$0leq left(n-1right)a_n-na_{n+1}Longleftrightarrow na_{n+1}leq left(n-1right)a_n$$

That means that $a_n$ is decreasing and bounded, so it does converges. Now define the sum

$$sum b_n = sum left(n-1right)a_n-na_{n+1}$$

Which is a telescop-sum and thus it converges

But that implies that the sum over $a_n$
$$sum a_n leq sum left(L-1right)a_n leq sum b_n$$
is convergent as well by the comparison test.

share|cite|improve this answer

edited Jan 8 '14 at 14:49

answered Jan 8 '14 at 14:22

user127.0.0.1user127.0.0.1

6,00162139

$endgroup$

add a comment | 

4

$begingroup$

The series even converges absoute, thus set $a_n:=left|a_nright|$

Now (if $n$ is big enough)
$$L = lim_{ntoinfty}nleft(1-frac{a_{n+1}}{a_n}right) Longleftrightarrow frac{a_{n+1}}{a_n}leqfrac{n-L}{n}$$

This is equivalent to
$$left(L-1right)a_n leq left(n-1right)a_n-na_{n+1}$$

Because of $L>1$ the left side is bigger than zero, so

$$0leq left(n-1right)a_n-na_{n+1}Longleftrightarrow na_{n+1}leq left(n-1right)a_n$$

That means that $a_n$ is decreasing and bounded, so it does converges. Now define the sum

$$sum b_n = sum left(n-1right)a_n-na_{n+1}$$

Which is a telescop-sum and thus it converges

But that implies that the sum over $a_n$
$$sum a_n leq sum left(L-1right)a_n leq sum b_n$$
is convergent as well by the comparison test.

share|cite|improve this answer

edited Jan 8 '14 at 14:49

answered Jan 8 '14 at 14:22

user127.0.0.1user127.0.0.1

6,00162139

$endgroup$

add a comment | 

$begingroup$

The series even converges absoute, thus set $a_n:=left|a_nright|$

Now (if $n$ is big enough)
$$L = lim_{ntoinfty}nleft(1-frac{a_{n+1}}{a_n}right) Longleftrightarrow frac{a_{n+1}}{a_n}leqfrac{n-L}{n}$$

This is equivalent to
$$left(L-1right)a_n leq left(n-1right)a_n-na_{n+1}$$

Because of $L>1$ the left side is bigger than zero, so

$$0leq left(n-1right)a_n-na_{n+1}Longleftrightarrow na_{n+1}leq left(n-1right)a_n$$

That means that $a_n$ is decreasing and bounded, so it does converges. Now define the sum

$$sum b_n = sum left(n-1right)a_n-na_{n+1}$$

Which is a telescop-sum and thus it converges

But that implies that the sum over $a_n$
$$sum a_n leq sum left(L-1right)a_n leq sum b_n$$
is convergent as well by the comparison test.

share|cite|improve this answer

edited Jan 8 '14 at 14:49

answered Jan 8 '14 at 14:22

user127.0.0.1user127.0.0.1

6,00162139

$endgroup$

share|cite|improve this answer

edited Jan 8 '14 at 14:49

answered Jan 8 '14 at 14:22

user127.0.0.1user127.0.0.1

6,00162139

answered Jan 8 '14 at 14:22

user127.0.0.1user127.0.0.1

6,00162139

answered Jan 8 '14 at 14:22

user127.0.0.1user127.0.0.1

6,00162139

1

$begingroup$

The following is a new argument for the convergence part. Assume that $L>1$. Choose $varepsilon > 0$ small enough such that
$$L-varepsilon > 1.$$
There exists some $1 ll N=N(varepsilon)$ such that
$$nBig( 1- frac {a_{n+1}}{a_n} Big) > L-varepsilon$$
for any $n geq N$, namely
$$frac {a_{n+1}}{a_n} < 1 - frac{L-varepsilon}n = frac{n-(L-varepsilon)}n$$
for any $n geq N$. Since
$L-varepsilon>1$, one can always choose $alpha >1$ in such a way that
$$frac{n-(L-varepsilon)}n < Big( frac{n-1}n Big)^alpha$$
for any $n gg 1$, say $n geq M > N$. Hence for large $n$, we obtain
$$ a_{n+1} leq Big( frac{n-1}n Big)^alpha a_n leq Big( frac{n-2}n Big)^alpha a_{n-1} leq cdots leq Big( frac{M-1}n Big)^alpha a_M.$$
Hence $sum a_n$ converges by the comparison test.

share|cite|improve this answer

answered Mar 13 at 6:34

QA NgôQA Ngô

5915

$endgroup$

add a comment | 

1

$begingroup$

The following is a new argument for the convergence part. Assume that $L>1$. Choose $varepsilon > 0$ small enough such that
$$L-varepsilon > 1.$$
There exists some $1 ll N=N(varepsilon)$ such that
$$nBig( 1- frac {a_{n+1}}{a_n} Big) > L-varepsilon$$
for any $n geq N$, namely
$$frac {a_{n+1}}{a_n} < 1 - frac{L-varepsilon}n = frac{n-(L-varepsilon)}n$$
for any $n geq N$. Since
$L-varepsilon>1$, one can always choose $alpha >1$ in such a way that
$$frac{n-(L-varepsilon)}n < Big( frac{n-1}n Big)^alpha$$
for any $n gg 1$, say $n geq M > N$. Hence for large $n$, we obtain
$$ a_{n+1} leq Big( frac{n-1}n Big)^alpha a_n leq Big( frac{n-2}n Big)^alpha a_{n-1} leq cdots leq Big( frac{M-1}n Big)^alpha a_M.$$
Hence $sum a_n$ converges by the comparison test.

share|cite|improve this answer

answered Mar 13 at 6:34

QA NgôQA Ngô

5915

$endgroup$

add a comment | 

$begingroup$

The following is a new argument for the convergence part. Assume that $L>1$. Choose $varepsilon > 0$ small enough such that
$$L-varepsilon > 1.$$
There exists some $1 ll N=N(varepsilon)$ such that
$$nBig( 1- frac {a_{n+1}}{a_n} Big) > L-varepsilon$$
for any $n geq N$, namely
$$frac {a_{n+1}}{a_n} < 1 - frac{L-varepsilon}n = frac{n-(L-varepsilon)}n$$
for any $n geq N$. Since
$L-varepsilon>1$, one can always choose $alpha >1$ in such a way that
$$frac{n-(L-varepsilon)}n < Big( frac{n-1}n Big)^alpha$$
for any $n gg 1$, say $n geq M > N$. Hence for large $n$, we obtain
$$ a_{n+1} leq Big( frac{n-1}n Big)^alpha a_n leq Big( frac{n-2}n Big)^alpha a_{n-1} leq cdots leq Big( frac{M-1}n Big)^alpha a_M.$$
Hence $sum a_n$ converges by the comparison test.

share|cite|improve this answer

answered Mar 13 at 6:34

QA NgôQA Ngô

5915

$endgroup$

share|cite|improve this answer

answered Mar 13 at 6:34

QA NgôQA Ngô

5915

answered Mar 13 at 6:34

QA NgôQA Ngô

5915

answered Mar 13 at 6:34

QA NgôQA Ngô

5915